Math 220-2
Homework #9 Solution (even numbers):
Section 5.2:
# 2 (4 points)
3 
5 3
5  
Ans: A  
,
A  I  
. The characteristic polynomial is

5   
3 5
 3
det( A  I )  (5   )(5   )  3  3  2  10  16  (  8)(  2)
So, the eigenvalues of A are 8 and 2.
# 16
Ans:
The determinant of a triangular matrix is the product of its diagonal entries:
0
0
0 
5  
 8
4
0
0 

det( A  I )  det
 (5   )( 4   )(1   ) 2
 0
7
1 
0 


5
2
1 
 1
The eigenvalues are 5, 1, 1, and -4.
Section 5.3:
# 2 (4 points)
 2  3
1 0 
,D  
, A  PDP 1 , and
Ans: P  


 3 5 
0 1 / 2
0 
5 3 4 1
P 1  
,D  

, and
3 2
0 1 / 16
A 4  PD 4 P 1 .we
compute
0  5 3 1  151
90 
 2  3 1
A4  
 






 3 5  0 1 / 16 3 2 16  225  134
# 8 (4 points)
Ans: Since A is triangular, its only eigenvalue is obviously 5.
0 1
For λ=5: A  5I  
. The equation (A-5I)x=0 amounts to x2 = 0, so x2 = 0
0 0
1 
with x1 free. The general solution is x1  . Since we cannot generate an
0 
eigenvector basis for R^2, A is not diagonalizable.
# 10
Ans:
To find the eigenvalues of A, compute its characteristic polynomial:
2   3 
det( A  I )  det 
 (2   )(1   )  3  4  2  3  10  (  5)(  2)

 4 1 
Thus the eigenvalues of A are 5 and -2.
 3 3 
For λ=5: A  5I  
 . The equation (A-5I)x = 0 amounts to x1-x2 = 0, so
 4  4
1
x1=x2 with x2 free. The general solution is x 2   , and a basis vector for the
1
1
eigenspace is v1    .
1
4 3
For λ=-2: A  2 I  
. The equation (A+2I)x = 0 amounts to 4x1+3x2 = 0,
4 3
  3 / 4
so x1 = (-3/4)x2 with x2 free. The general solution is x 2 
, and a nice basis
 1 
 3
vector for the eigenspace is v 2    .
4
1  3
5 0 
v2   
. Then set D  

, where
1 4 
0  2 
the eigenvalues in D correspond to v1 and v2 respectively.
From v1 and v2 construct P  v1
# 22
Ans:
a). False. The n eigenvectors must be linearly independent. See the
Diagonalization Theorem.
b). False. The matrix in Example 3 is diagonalizable, but it has only 2 distinct
eigenvalues.
c). True. This follows from AP=PD and formulas (1) and (2) in the proof of the
Diagonalization Theorem.
d). False. See Example 4. The matrix there is invertible because 0 is not an
eigenvalue, but the matrix is not diagonalizable.
Section 6.1:
#2
Ans:
3
Since w    1
 5
and
6
x   2 ,
 3 
w  w  3 2  (1) 2  (5) 2  35
x  w  6  3  (2)( 1)  3(5)  5
xw
5 1

 .
w  w 35 7
#10 (4 points)
Ans: A unit vector in the direction of the given vector is
  6
 6  6 / 61
1
 4   1  4    4 / 61 


 
61   
(6) 2  4 2  (3) 2   3




3

3
/
61
 
  

#16
Ans:
#20
Ans:
Since u*v = 12(2)+(3)(-3)+(-5)(3)=0, u and v are orthogonal.
Since u*u is the sum of the squares of the entries in u, u*u ≥ 0. The sum of
squares of numbers is zero if and only if all the numbers are themselves zero.
Note: Completeness (4 points)